# Max amplitude of a diving board that will not toss a pebble off?

In summary: Of course it is. But, the contact with the board can be visualized with a free body diagram. Just stating that the acceleration of the board must be greater than the acceleration due to gravity is not providing a direct statement and connection regarding contact or not with the...
@malawi_glenn so from now on... looking at oscillations from other system, the thing only thing they have in common w/ the mass-spring system are the oscillation equations (that is the kinematics, nothing about their dynamics (besides the accelerations, right? )
just want to make sure. TIA

malawi_glenn said:
and the normal force on the pebble is 0 then because... (do not mention Newtons 2nd law)
If the board is acceleration down faster than the pebble then contact is lost and therefore no force.

Well, the fact that you get sinusodial motion does come from the equation of motion ## \ddot x - \frac{k}{m}x = 0##... but you do not need to worry about those details here, you just need to know that the acceleration of the pebble when it has contact with the board is ##\omega^2 x## in magnitude.

malawi_glenn said:
Well, the fact that you get sinusodial motion does come from the equation of motion ## \ddot x - \frac{k}{m}x = 0##... but you do not need to worry about those details here, you just need to know that the acceleration of the pebble when it has contact with the board is ##\omega^2 x## in magnitude.
got it that's indeed just the derivation the 2nd order ODE, I was looking at it from the kinetics POV only.

malawi_glenn said:
The force of gravity on the board is not acting on the pebble, that is the point of FBD. You draw the forces acting on the object of interest.
To add to that, the board makes simple harmonic motion relative to its rest position, not relative to the position where it has no internal stresses. This will effectively cancel out the weight term from the equations of motion for the end if the board.

Since you have the result already, let me just mention how my argumentation would go:

As long as the boardâ€™s acceleration downwards is smaller than g, the pebble will stay (because it cannot fall theough the board). The limiting case is a = g. The amplitude of the boardâ€™s acceleration is ##A\omega^2## which means that the max downwards acceleration is precisely this. Thus, the limiting case becomes ##g = A\omega^2##. Solve for ##A##.

Delta2

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